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An airplane is flying through a thundercloud at a height of 2010 m. (A very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If there is a charge concentration of 46.9 C at height 4090 m within the cloud and −59.7 C at height 570 m, what is the magnitude of the electric field E at the aircraft? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of V/m.

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Answer:

The value is
E = 3.5619 *10^(5) V/m

Step-by-step explanation:

From the question we are told that

The height of the airplane is
h = 2010 \ m

The first charge concentration is
Q = 46.9 \ C

The height of the first charge concentration is
h_1 = 4090 \ m

The second charge concentration is
Q_2 = −59.7 \ C

The height of the second charge concentration is
h_1 = 570 \ m

The electric field due to the first charge concentration is


E_1 = (k * |Q_1|)/((h_1 -h)^2)

Here k is the Coulomb constant given in the question


E_1 = (8.98755*10^9 * |46.9|)/((4090 - 2010)^2)


E_1 = 9.7429 *10^(4) \ V/m

The electric field due to the second charge concentration is


E_2 = (k * |-59.7|)/((h -h_2)^2)


E_2 = (8.98755*10^9 * |-59.7|)/((2010 - 570)^2)


E_2 = 2.5876 *10^(5) V/m

Generally the superposition principle can be applied in this question as follows


E = E_1 + E_2

=>
E = 9.7429 *10^(4)+ 2.5876 *10^(5)

=>
E = 3.5619 *10^(5) V/m

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